I understand what this means but I just don't understand what "generated by $C$" means.

Similarly, I am given an example of the Borel $\sigma$- algebra as $\sigma(T)$ where $T = ${open sets of $\mathbb{R}$}

So a Borel $\sigma$- algebra is equal to a $\sigma$- algebra generated by all the open sets of $\mathbb{R}$

Can someone please explain what the word "generated by" means?

I know a $\sigma$- algebra is a collection of subsets of the power set $2^{\Omega}$ where $\Omega$ is any set. So does this mean that if a $\sigma$- algebra is generated by something else, that something else is just the set $\Omega$?

$\begingroup$I think of "The sigma-algebra generated by $C$" as "everything reachable from $C$ through unions, intersections, and compliments"$\endgroup$
– JoeFeb 25 at 18:52

$\begingroup$"everything reachable from $C$" meaning that whatever is in the sigma algebra is whatever is reachable from $C$ through unions, intersections and complements?$\endgroup$
– user477465Feb 25 at 18:58

1

$\begingroup$I wouldn't recommend thinking about it like that. Since infinite unions may occur, such a "construction" of the generated sets is impossible in general. Also, this idea may lead you to seriously underestimate the complexity of a $\sigma$-algebra that may be generated by a very simple generator.$\endgroup$
– Mars PlasticFeb 25 at 19:01

$\begingroup$@MarsPlastic that's true, but I still think it's an "easy" way to think about the rough idea. Obviously there's the caveat that you can't formally construct the $\sigma(C)$ as "all finite sequences of elements of $C, \cap, \cup,$ and compliments"$\endgroup$
– JoeFeb 25 at 19:10

$\begingroup$@Joe You are right in that this may help to get a basic idea, but one should be very careful. This interpretation is more appropriate for an algebra.$\endgroup$
– Mars PlasticFeb 25 at 19:14

2 Answers
2

You can give meaning to the "smallest" $\sigma$-algebra containing $C\subset2^\Omega$ by first noting that any intersection of $\sigma$-algebras on $\Omega$ is again a $\sigma$-algebra on $\Omega$ and then setting

$\begingroup$interesting, i did have some confusion on what the difference was between $C$ and $A$. is $C \subset A$? is $C$ also a $\sigma$- algebra?$\endgroup$
– user477465Feb 25 at 19:21

$\begingroup$I edited my answer slightly which should take care of your first question. Concerning the second one: $C$ does not have to be a $\sigma$-algebra, as the whole point is to find the smallest $\sigma$-algebra containing $C$. If $C$ was one already, we'd obviously have $\sigma(C)=C$.$\endgroup$
– Mars PlasticFeb 25 at 19:27

$\begingroup$thanks. so if $A$ is a $\sigma$- algebra, what would it mean for a subset of A to not be a $\sigma$- algebra? If $\Omega$ is $\{1,2,4\}$ for example and $A =\{\{1\},\{2\}\}$, then can $C$ be $\{\{1\}\}$ it can't be $\{\{1\},\{2\},\{4\}\}$ right? In what instance would $C$ not be a subset of the power set? is it because my example was a finite set?$\endgroup$
– user477465Feb 25 at 19:38

$\begingroup$I'm not sure if I understand your question. In your case, $A=\{\{1\},\{2\}\}$ is not a $\sigma$-algebra. And $C$ has to be a subset of $2^\Omega$ for the initial question to even make sense.$\endgroup$
– Mars PlasticFeb 26 at 11:30

An intersection of $\sigma$-algebras is a $\sigma$-algebra, similar to the fact that an intersection of subgroups is a subgroup. This fact gives us that there is a unique minimal $\sigma$-algebra containing $C$ arising from a collection $C$, such that it generates the $\sigma$-algebra in that sense.

Which is again similar to what it means for a subgroup to be generated by a set, for a topology to be generated by a collection of sets and so on.